Question 1
If x3 + y3 = 9 and x + y = 3, then the value of x4+y4 is,
 A 21 B 0 C 17 D 25
Algebra
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Question 1 Explanation:
```x3+y3 = (x + y) × (x2 − xy + y2)

Putting given values of x3+y3 and (x + y)
9 = 3 × ((x+y)2 − 3xy)
= 3 × (9 − 3xy)
= 27 − 9xy

9xy = 18
xy = 2

x4 + y4 = (x2 + y2)2 - 2x2y2
= (x2 + y2)2 - 2*4
[Putting value of xy]
= ((x + y)2 - 2xy)2 - 2*4
[Putting values of (x+y) and xy]
= (9 - 4)2 - 2*4
= 17```
 Question 2
If x+1/2x = 2, find the value of 8x3+1/x3
 A 40 B 20 C 28 D 35
Algebra
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 Question 3
For any real number x the maximum value of 4−6x−x2 is at x=,
 A 4 B 6 C -3 D 3
Algebra
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Question 3 Explanation:
Differentiate and equate to 0 6+2x =0 x=−3
 Question 4
If 5√x +12√x =13√x then value of x is,
 A 2 B 1 C 3 D 4
Algebra
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Question 4 Explanation:
say x=1
17!=13

for, x=2 and 3 also not possible
x=4
52+122=132
169=169
 Question 5
 A 0 B -2 C 2 D 4
Algebra
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Question 5 Explanation:
 Question 6
 A 0 B 6 C -4 D 4
Algebra
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Question 6 Explanation:
 Question 7
 A 8 B 2 C 0 D 6
Algebra
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Question 7 Explanation:
 Question 8
 A 7/4 B 9/4 C 5/4 D 3/4
Algebra
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Question 8 Explanation:
 Question 9
 A 6 B 2 C 3 D 0
Algebra
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Question 9 Explanation:
Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab + ac
(ab + bc + ca)/[(a+b+c)(abc)]
as, ab+bc+ca=0
= 0
 Question 10
 A A B B C C D D E E
Algebra    Data Sufficiency
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 Question 11
4/15 of 5/7 of a number is greater than 4/9 of 2/5 of the same number by 8. What is half of that number?
 A 630 B 63 C 315 D 105
Algebra
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Question 11 Explanation:
Let the number be x. Then, 4/15 of 5x/7 = 4/9 of 2x/5 + 8 4/15 * 5x/7 = 4/9 * 2x/5 + 8 4x/21 = 8x/45 + 8 x/21 = 2x/45 + 1 x/21 = (2x + 45)/45 45x = 21(2x + 45) 45x = 42x + 45*21 3x = 45*21 x = 15*21 = 315.
 Question 12
If 2x/(1 + 1/(1 + x/(1 - x))) = 1, then find the value of x.
 A 1/3 B 2/3 C 1/2 D 3/2
Algebra
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Question 12 Explanation:
We have 2x/(1 + 1/(1 + x/(1-x))) = 1 2x/(1 + 1/((1-x+x)/(1-x))) = 1 2x/(1 + 1/(1/(1-x))) = 1 2x/(1 + 1-x) = 1 2x/(2-x) = 1 2x = 2-x 3x = 2 x = 2/3.
 Question 13
Find the value of (1-1/3)(1-1/4)(1-1/5)...(1-1/100).
 A 1/10 B 1/50 C 1/20 D 1/100
Algebra
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Question 13 Explanation:
We have (1-1/3)(1-1/4)(1-1/5)...(1-1/100) = (2/3)(3/4)(4/5)...(99/100) = 2/100 = 1/50.
 Question 14
A zoo has some peacocks and some horses. If the number of heads be 48 and the number of feet be 140, then the number of peacocks will be:
 A 23 B 25 C 26 D 27
Algebra
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Question 14 Explanation:
Let there be x peacocks and y horses. Then, x + y = 48 (equation 1) and 2x + 4y = 140 x + 2y = 70 Substituting the value of x from equation 1, we get 48 -y + 2y = 70 y = 22 Then, x = 48 - 22 = 26.
 Question 15
A total of 324 coins of 20 paise and 25 paise make a sum of Rs 71. The number of 25-paise coins is:
 A 104 B 120 C 128 D 124
Algebra
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Question 15 Explanation:
Let the number of 20-paise coins be x and the number of 25-paise coins be y. Then, x + y = 324 (equation 1) And, x/5 + y/4 = 71 4x + 5y = 71*20 (Multiplying by 20 on both sides) Substituting the value of x from equation 1, we get 4(324-y) + 5y = 71*20 1296 - 4y + 5y = 1420 y = 124.
 Question 16
 A 1 B 2 C 3 D 0
Algebra    GATE 2017 Mock
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Question 16 Explanation:

Correction: Instead of R3 -> R3 + 3R it should be R3 -> R3 + 3R1 while applying the first set of rules.
There are 16 questions to complete.

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